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        <span title="转载" class="icon-repost"><i class="fa-solid fa-share fa-fw" aria-hidden="true"></i></span><span>1806.还原排列的最少操作步数</span>
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      <div class="post-meta-line"><span title=2023-01-09&#32;23:02:06><i class="fa-regular fa-calendar-alt fa-fw" aria-hidden="true"></i>&nbsp;<time datetime="2023-01-09">2023-01-09</time></span>&nbsp;<span><i class="fa-solid fa-pencil-alt fa-fw" aria-hidden="true"></i> 约 532 字</span>&nbsp;<span><i class="fa-regular fa-clock fa-fw" aria-hidden="true"></i> 预计阅读 2 分钟</span>&nbsp;<span id="/posts/1806.%E8%BF%98%E5%8E%9F%E6%8E%92%E5%88%97%E7%9A%84%E6%9C%80%E5%B0%91%E6%93%8D%E4%BD%9C%E6%AD%A5%E6%95%B0/" class="leancloud_visitors comment-visitors" data-flag-title="1806.还原排列的最少操作步数">
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        <li><a href="#题目描述">题目描述</a></li>
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      </div><div class="content" id="content"><h2 id="题目描述">题目描述</h2>
<p>给你一个偶数 <code>n</code> ，已知存在一个长度为 <code>n</code> 的排列 <code>perm</code> ，其中 <code>perm[i] == i</code>（下标 <strong>从 0 开始</strong> 计数）。</p>
<p>一步操作中，你将创建一个新数组 <code>arr</code> ，对于每个 <code>i</code> ：</p>
<ul>
<li>如果 <code>i % 2 == 0</code> ，那么 <code>arr[i] = perm[i / 2]</code></li>
<li>如果 <code>i % 2 == 1</code> ，那么 <code>arr[i] = perm[n / 2 + (i - 1) / 2]</code></li>
</ul>
<p>然后将 <code>arr</code> 赋值给 <code>perm</code> 。</p>
<p>要想使 <code>perm</code> 回到排列初始值，至少需要执行多少步操作？返回最小的 <strong>非零</strong> 操作步数。</p>
<p><strong>示例 1：</strong></p>
<div class="highlight"><div class="chroma">
<table class="lntable"><tr><td class="lntd">
<pre tabindex="0" class="chroma"><code><span class="lnt">1
</span><span class="lnt">2
</span><span class="lnt">3
</span><span class="lnt">4
</span><span class="lnt">5
</span></code></pre></td>
<td class="lntd">
<pre tabindex="0" class="chroma"><code class="language-fallback" data-lang="fallback"><span class="line"><span class="cl">输入：n = 2
</span></span><span class="line"><span class="cl">输出：1
</span></span><span class="line"><span class="cl">解释：最初，perm = [0,1]
</span></span><span class="line"><span class="cl">第 1 步操作后，perm = [0,1]
</span></span><span class="line"><span class="cl">所以，仅需执行 1 步操作
</span></span></code></pre></td></tr></table>
</div>
</div><p><strong>示例 2：</strong></p>
<div class="highlight"><div class="chroma">
<table class="lntable"><tr><td class="lntd">
<pre tabindex="0" class="chroma"><code><span class="lnt">1
</span><span class="lnt">2
</span><span class="lnt">3
</span><span class="lnt">4
</span><span class="lnt">5
</span><span class="lnt">6
</span></code></pre></td>
<td class="lntd">
<pre tabindex="0" class="chroma"><code class="language-fallback" data-lang="fallback"><span class="line"><span class="cl">输入：n = 4
</span></span><span class="line"><span class="cl">输出：2
</span></span><span class="line"><span class="cl">解释：最初，perm = [0,1,2,3]
</span></span><span class="line"><span class="cl">第 1 步操作后，perm = [0,2,1,3]
</span></span><span class="line"><span class="cl">第 2 步操作后，perm = [0,1,2,3]
</span></span><span class="line"><span class="cl">所以，仅需执行 2 步操作
</span></span></code></pre></td></tr></table>
</div>
</div><p><strong>示例 3：</strong></p>
<div class="highlight"><div class="chroma">
<table class="lntable"><tr><td class="lntd">
<pre tabindex="0" class="chroma"><code><span class="lnt">1
</span><span class="lnt">2
</span></code></pre></td>
<td class="lntd">
<pre tabindex="0" class="chroma"><code class="language-fallback" data-lang="fallback"><span class="line"><span class="cl">输入：n = 6
</span></span><span class="line"><span class="cl">输出：4
</span></span></code></pre></td></tr></table>
</div>
</div><p><strong>提示：</strong></p>
<ul>
<li><code>2 &lt;= n &lt;= 1000</code></li>
<li><code>n</code> 是一个偶数</li>
</ul>
<h2 id="分析">分析</h2>
<ol>
<li>
<p>当n=4时，<u>0</u>12<u>3</u> -&gt; 0213 -0123, 我们发现第一位和最后一位位置不变，只有中间的数字在动</p>
</li>
<li>
<p>当n=6时，012345 -&gt; 024135 -&gt; 043215 -&gt; 031425 -&gt; 012345, 和上面是一样的规律</p>
</li>
<li>
<p>同时我们发现，当1回到原来的位置时，所有的数字都回到原来的位置上了</p>
</li>
</ol>
<p>所以，我可以数字 <code>1</code> 作为锚点，当数字 <code>1</code> 回到原点则所有数字都回到原位</p>
<h2 id="解法">解法</h2>
<div class="highlight"><div class="chroma">
<table class="lntable"><tr><td class="lntd">
<pre tabindex="0" class="chroma"><code><span class="lnt"> 1
</span><span class="lnt"> 2
</span><span class="lnt"> 3
</span><span class="lnt"> 4
</span><span class="lnt"> 5
</span><span class="lnt"> 6
</span><span class="lnt"> 7
</span><span class="lnt"> 8
</span><span class="lnt"> 9
</span><span class="lnt">10
</span><span class="lnt">11
</span><span class="lnt">12
</span><span class="lnt">13
</span><span class="lnt">14
</span><span class="lnt">15
</span><span class="lnt">16
</span><span class="lnt">17
</span><span class="lnt">18
</span><span class="lnt">19
</span><span class="lnt">20
</span></code></pre></td>
<td class="lntd">
<pre tabindex="0" class="chroma"><code class="language-go" data-lang="go"><span class="line"><span class="cl"><span class="kd">func</span> <span class="nf">reinitializePermutation</span><span class="p">(</span><span class="nx">n</span> <span class="kt">int</span><span class="p">)</span> <span class="kt">int</span> <span class="p">{</span>
</span></span><span class="line"><span class="cl">  <span class="c1">// 用1作为锚点
</span></span></span><span class="line"><span class="cl"><span class="c1"></span>	<span class="nx">tmp</span> <span class="o">:=</span> <span class="mi">1</span>
</span></span><span class="line"><span class="cl">  <span class="c1">// 记录操作数
</span></span></span><span class="line"><span class="cl"><span class="c1"></span>	<span class="nx">cnt</span> <span class="o">:=</span> <span class="mi">0</span>
</span></span><span class="line"><span class="cl">	<span class="k">for</span> <span class="p">{</span>
</span></span><span class="line"><span class="cl">		<span class="k">if</span> <span class="nx">tmp</span><span class="o">%</span><span class="mi">2</span> <span class="o">==</span> <span class="mi">1</span> <span class="p">{</span>
</span></span><span class="line"><span class="cl">			<span class="nx">tmp</span> <span class="p">=</span> <span class="nx">n</span><span class="o">/</span><span class="mi">2</span> <span class="o">+</span> <span class="p">(</span><span class="nx">tmp</span><span class="o">-</span><span class="mi">1</span><span class="p">)</span><span class="o">/</span><span class="mi">2</span>
</span></span><span class="line"><span class="cl">		<span class="p">}</span> <span class="k">else</span> <span class="p">{</span>
</span></span><span class="line"><span class="cl">			<span class="nx">tmp</span> <span class="o">/=</span> <span class="mi">2</span>
</span></span><span class="line"><span class="cl">		<span class="p">}</span>
</span></span><span class="line"><span class="cl">    <span class="c1">// 操作+1
</span></span></span><span class="line"><span class="cl"><span class="c1"></span>		<span class="nx">cnt</span><span class="o">++</span>
</span></span><span class="line"><span class="cl">    <span class="c1">// 数字1回到原点，则退出循环
</span></span></span><span class="line"><span class="cl"><span class="c1"></span>		<span class="k">if</span> <span class="nx">tmp</span> <span class="o">==</span> <span class="mi">1</span> <span class="p">{</span>
</span></span><span class="line"><span class="cl">			<span class="k">break</span>
</span></span><span class="line"><span class="cl">		<span class="p">}</span>
</span></span><span class="line"><span class="cl">	<span class="p">}</span>
</span></span><span class="line"><span class="cl">	<span class="k">return</span> <span class="nx">cnt</span>
</span></span><span class="line"><span class="cl"><span class="p">}</span>
</span></span></code></pre></td></tr></table>
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